1 - 2 - k Game

题目描述
Alice and Bob play a game. There is a paper strip which is divided into n + 1 cells numbered from left to right starting from 0. There is a chip placed in the n-th cell (the last one).

Players take turns, Alice is first. Each player during his or her turn has to move the chip 1, 2 or k cells to the left (so, if the chip is currently in the cell i, the player can move it into cell i - 1, i - 2 or i - k). The chip should not leave the borders of the paper strip: it is impossible, for example, to move it k cells to the left if the current cell has number i < k. The player who can't make a move loses the game.

Who wins if both participants play optimally?

Alice and Bob would like to play several games, so you should determine the winner in each game.

输入
The first line contains the single integer T (1 ≤ T ≤ 100) — the number of games. Next T lines contain one game per line. All games are independent.

Each of the next T lines contains two integers n and k (0 ≤ n ≤ 10^9, 3 ≤ k ≤ 10^9) — the length of the strip and the constant denoting the third move, respectively.

输出
For each game, print Alice if Alice wins this game and Bob otherwise.

样例输入
4
0 3
3 3
3 4
4 4

样例输出
Bob
Alice
Bob
Alice

题意：有一个东西放在n这个位置 每次可以选择向左移动 1 、2 或者 k 步 不能小于 0 ，不能移动则输。

#include <cctype>
#include <map>
#include <set>
#include <vector>
#include <string>
#include <stack>
#include <queue>

#define lowbit(x) x & (-x)
#define lson left,mid,root << 1
#define rson mid+1,right,root << 1 | 1

using namespace std;

typedef pair<int, int> PII;
typedef long long ll;
typedef unsigned long long ull;
const ll mod=1e9+7;
const ll inf=0x3f3f3f3f;
const double eps=1e-8;
const int maxn=1e5+100;

int main()
{
    int t;
    ll n,k;
    scanf("%d",&t);
    while(t --)
        {
            scanf("%lld%lld",&n,&k);
            if(n == 0)
                {
                puts("Bob");
                continue;
                }
            if(k % 3)
                {
                    if(n % 3)
                        puts("Alice");
                    else
                        puts("Bob");
                }
            else
                {
                int t = n % (k + 1);
                if(t % 3 || t == k)
                    puts("Alice");
                else
                    puts("Bob");
                }
        }
    return 0;
}